The sodium-22 nuclide radioactively decays by positron emission. Write a balanced nuclear chemical equation that describes this process.

Sure, here is the balanced nuclear chemical equation for the radioactive decay of Sodium-22 (Na-22) by positron emission:

Na-22 → Ne-22 + e+

Here's the step-by-step process:

1. Identify the initial nuclide, which is Sodium-22 (Na-22). In nuclear notation, this is written as 22/11 Na (where 22 is the mass number and 11 is the atomic number).

2. Identify the type of decay, which is positron emission. In positron emission, a proton in the nucleus is converted into a neutron, a positron, and a neutrino. The positron (e+) is the particle that is emitted, and it has a charge of +1 and a mass number of 0.

3. Write the final nuclide. Since a proton is lost in the process, the atomic number decreases by 1, resulting in Neon-22 (Ne-22). In nuclear notation, this is written as 22/10 Ne.

4. Balance the equation. The sum of the mass numbers on the left side of the equation must equal the sum of the mass numbers on the right side of the equation. Similarly, the sum of the atomic numbers on the left side of the equation must equal the sum of the atomic numbers on the right side of the equation.

So, the balanced nuclear chemical equation is:

22/11 Na → 22/10 Ne + 0/+1 e

This equation shows that Sodium-22 (Na-22) decays into Neon-22 (Ne-22) by emitting a positron (e+).